4 Answers
4

All of the answers given so far actually work without change when $W$ and $V$ are not necessarily finite-dimensional. To be completely explicit, notice that any basis $\mathcal{B}$ of $W$ is a linearly independent subset of $V$, and Zorn's lemma implies there is a maximal linearly independent subset $\mathcal{B}'$ of $V$ containing $\mathcal{B}$. It is not hard to show that $\mathcal{B}'$ is a basis of $V$, so we can construct a linear map $L : V \to V$ that sends every basis vector in $\mathcal{B}$ to $0$ while sending any basis vector in $\mathcal{B}' \setminus \mathcal{B}$ to itself. By construction, we have $\ker L = W$.

Assuming $V$ is finite dimensional consider a basis $\{v_1,v_2\cdots v_k\cdots v_n\}$ for $V$ where $\{v_1,v_2\cdots v_k\}$ is a basis for $W$. Now let $L$ send each vector in $\{v_i:1\le i\le k\}$ to $0$ and the vectors $v_{k+1},\cdots v_n$ to themselves. This gives you the requisite transformation as any $v\in V$ may be expressed as $v=\sum_{i=1}^n \alpha_i v_i$ and you can define $L(v)=\sum_{i=1}^n \alpha_i L(v_i)$.

I'm assuming that your vector spaces are finite-dimensional. So, you can take a base of W and then extend it to a base of V. Then remember that a linear map is completely defined by it's value on a base.

Let $\{e_1,\ldots,e_k\}$ be a base of $W$. Complete it to a base $\{e_1,\ldots,e_k,e_{k+1},\ldots,e_n\}$ of $V$. Define $Le_i=0$ if $1 \leq i \leq k$, and $Le_i=e_i$ if $k+1 \leq i \leq n$. You can extend by linearity this map.